Finding the Speed of a Box Tied to a Propeller on a Spring

[bodensee9]

Hello:
I was wondering if someone could check if the following is correct: A
box of mass M is tied to a propeller with an upward force of F. Ignore
weight of propeller. The box is placed on a vertical spring that is initially
at equilibrium and then the propeller is ignited. The spring constant is K.
Find speed after the spring has moved by a distance of X. So would I do:
K_{initial} = 0; U_{initial} = 0; U_{spring} = 0 initially. Kf = 1/2*mv^2
; U_{final} = MgX;U_{spring} = 1/2*kX^2.
And the total work done by the force F is F.x, or at the distance X, FX.
So since the total work done is equal to the change in mechanical energy, so
I write: FX = 1/2*mv^2 +MgX + 1/2*KX^2 and solve for v? Thank you.

 

[LowlyPion]

I see that the spring is not compressed at the start. So yes, the equation as you have written it looks right.

 

[bodensee9]

Hello:
Oh right, thanks!